What is the value of the likelihood ratio test statistic for a given poisson distribution?

In past weeks, the average number of life insurance policies sold per week was $$θ = 3$$. However, this week he has sold 6 life insurance policies. Based on this single observation of $$x = 6$$, what is the value of the likelihood-ratio test statistic for testing the hypothesis that the salesman has suddenly seen a boost in his job, i.e. testing $$H_0 = \theta = 3$$ , $$H_1 = \theta > 3$$.

For this, I used the likelihood ratio for poisson distribution, giving me $$\frac{\frac{3^6e^{-3}}{6!}}{\frac{6^6e^{-6}}{6!}}$$ This gives 0.3138365. Is this the right answer?

• Your fraction reduces to $3^3/6^6 = .5^6= 0.015625.$ – BruceET Apr 29 '19 at 19:52
• @BruceET - thought the fraction was probably intended to be $\frac{\frac{3^6 e^{-3}}{6!}}{\frac{6^6e^{-6}}{6!}} =\dfrac{\exp(3)}{2^6}$ – Henry Apr 29 '19 at 20:01
• Yes that was the intended fraction, just typed it wrong. Thank you for clarifying! – martinhynesone Apr 29 '19 at 20:15

You will want to reject $$H_0: \theta = 3$$ against $$H_a: \theta > 3)$$ for large values of your single observation $$X.$$ You can't test at exactly the 5% level because of the discreteness of the Poisson distribution.

Under $$H_0,$$ you have $$P(X \ge 7) = .0335,$$ but $$P(X \ge 6) = .0834.$$ Exact Poisson computations from R (where ppois is a Poisson CDF and qpois is the quantile function):

qpois(.95, 3)
[1] 6
1 - ppois(6,3)
[1] 0.03350854  # 1 - P(X <= 6) = P(X >= 7)
1 - ppois(5,3)
[1] 0.08391794


So for test at level $$\alpha =3.335,$$ you should reject if $$X \ge 7.$$ Because you observe $$X = 6,$$ the P-value is $$P(X \ge 6\,|\,\theta = 3) = 0.834.$$

In the plot below, $$\alpha$$ is the sum of the heights of the bars to the right of the vertical dotted line.

For such a small values of $$\theta,$$ depending on the desired accuracy, it may no bet advisable to use a normal approximation. The best-fitting normal curve is shown in blue.

However, using the normal approximation (with continuity correction), the P-value is 7.4%, so you would not reject at the 5% level.

1 - pnorm(5.5, 3, sqrt(3))
[1] 0.07445734


Addendum: Now that you have the correct likelihood ratio function $$\lambda(x) = (3/x)^xe^{x-3}$$ for $$x \ge 3 \; (1$$ for $$x < 3),$$ let's use R to plot it. Then we can see that $$\lambda(x)$$ is small (leading to rejection) for large values of $$x$$ as mentioned at the start.

x = 3:10;  like = (3/x)^x*exp(x - 3)
plot(x, like, pch=20, main="Likelihood Ratio")
abline(h = 0, col="green2")